$ \int_0^1 \int_0^2 dy \, dx + \int_1^2 \int_{2(x - 1)}^2 dy \, dx$ Switch the bounds of the double integral. Choose 1 answer: Choose 1 answer: (Choice A) A $ \int_0^4 \int_{1 + y/2}^2 dx \, dy$ (Choice B) B $ \int_0^2 \int_0^{1 + y/2} dx \, dy$ (Choice C) C $ \int_2^4 \int_1^{1 + y/2} dx \, dy$ (Choice D) D $ \int_0^2 \int_{1 + y/2}^2 dx \, dy$
Solution: The first step whenever we want to switch bounds is to sketch the region of integration that we're given. We are given two regions in this case. The first has $0 < x < 1$ and $0 < y < 2$. The second has $1 < x < 2$ and $2(x - 1) < y < 2$. Therefore: ${2}$ ${4}$ ${2}$ ${4}$ $y$ $x$ Because we're switching bounds to $dx \, dy$, we need to start with numeric bounds for $y$. We see that $0 < y < 2$. Then we can define $x$ in terms of $y$. Thus, $0 < x < 1 + \dfrac{y}{2}$. In conclusion, the double integral after switching bounds is: $ \int_0^2 \int_0^{1 + y/2} dx \, dy$